http://kishoremaths.in/ ALGEBRA POLYNOMIALS A polynomial in x is an expression of the form p(x) = a 0 , + a 1 x + a 2 x 2 + ………………. + a n x n Where a 0 , a 1 , a 2 …………. a n are real numbers and n is a non-negative integer and a n 0. A polynomial having only one term is called monomial. A polynomial having two terms is called binominal. A polynomial having three terms is called trinomial. Degree of polynomial: The highest power of x in a polynomial is called degree of a polynomial. A polynomial is said to be linear if it is of degree 1. A polynomial is said to be quadratic if it is of degree 2. A polynomial is said to be cubic if it is of degree 3. A polynomial is said to be biquadratic if it is of degree 4. Sum and difference of two polynomial Let p(x) and q(x) be two polynomials p(x) = x 4 + 3x 3 + 2x 2 + 6x + 9 q(x) = 2x 3 + 3x 2 + 6x + 3 The sum of two polynomial gives a polynomial Here p(x) + q(x)=x 4 + 5x 3 + x 2 + 12x + 12 Similarly we can find difference of two polynomials p(x) – q(x)= x 4 + x 3 – 5x 2 + 6 Multiplication of two polynomial p(x)q(x) = (x 4 + 3x 3 – 2x 2 + 6x + 9)( 2x 3 + 3x 2 + 6x + 3) Division of a polynomial by a polynomial- If we divide p(x) by g(x) and we get quotient q(x) and remainder r(x) then we can write p(x) = g(x) q(x) + r(x) degree r(x) < degree g(x) Illustration 1: Divide p(x) by g(x), where p(x) = x 4 + 1 and g(x) = x + 1 Solution: x 3 – x 2 + x – 1 x + 1 x 4 + 1 x 4 + x 3 –x 3 + 1 –x 3 – x 2 x 2 + 1 x 2 + x –x + 1 –x – 1 2 Here, the quotient q(x) = x 3 – x 2 + x – 1 and the remainder r(x) = 2 We write x 4 + 1 = (x 3 – x 2 + x – 1) (x + 1) + 2.
Transcript
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ALGEBRA
POLYNOMIALS
A polynomial in x is an expression of the form
p(x) = a0, + a1x + a2x2 + ………………. + anx
n
Where a0, a1, a2…………. an are real numbers and n is a non-negative integer and an 0.
A polynomial having only one term is called monomial.
A polynomial having two terms is called binominal.
A polynomial having three terms is called trinomial.
Degree of polynomial: The highest power of x in a polynomial is called degree of a polynomial.
A polynomial is said to be linear if it is of degree 1.
A polynomial is said to be quadratic if it is of degree 2.
A polynomial is said to be cubic if it is of degree 3.
A polynomial is said to be biquadratic if it is of degree 4.
Sum and difference of two polynomial
Let p(x) and q(x) be two polynomials
p(x) = x4 + 3x
3 + 2x
2 + 6x + 9
q(x) = 2x3 + 3x
2 + 6x
+ 3
The sum of two polynomial gives a polynomial
Here p(x) + q(x)=x4 + 5x
3 + x
2 + 12x + 12
Similarly we can find difference of two polynomials
p(x) – q(x)= x4 + x
3 – 5x
2 + 6
Multiplication of two polynomial
p(x)q(x) = (x4 + 3x
3 – 2x
2 + 6x + 9)( 2x
3 + 3x
2 + 6x
+ 3)
Division of a polynomial by a polynomial- If we divide p(x) by g(x) and we get quotient q(x) and remainder
r(x) then we can write
p(x) = g(x) q(x) + r(x) degree r(x) < degree g(x)
Illustration 1: Divide p(x) by g(x), where p(x) = x4 + 1 and g(x) = x + 1
Solution: x3 – x
2 + x – 1
x + 1 x4 + 1
x4 + x
3
–x3 + 1
–x3 – x
2
x2 + 1
x2 + x
–x + 1
–x – 1
2
Here, the quotient q(x) = x3 – x
2 + x – 1 and the remainder r(x) = 2
We write x4 + 1 = (x
3 – x
2 + x – 1) (x + 1) + 2.
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Note: Notice that the degree of g(x) is less than the degree of p(x). Therefore it is always possible to divide a
polynomial of higher degree by a polynomial of lower degree (in the same variable). The process above
stops as soon as the remainder is zero or the degree of the remainder becomes smaller than that of the
divisor
Illustration 2: Let p(x) = x4 + 2x
3 – 3x
2 + x – 1. Find the remainder when p(x) is divided by x – 2.
Solution: x3 + 4 x
2 + 5x + 11
x – 2 x4 + 2x
3 – 3x
2 + x – 1
x4 – 2x
3
4x3 – 3x
2 + x – 1
4x3 – 8x
2
5x2 + x – 1
5x2 –10x
11x – 1
11x – 22
21
Remainder is 21.
In the above example let us evaluate p(2). We get
p(2) = 24 + 2(2
3) – 3x
2 + 2 – 1
= 16 + 16 – 12 + 2 – 1 = 21.
Thus we find that p(2) is equal to the remainder when p(x), is divided by x – 2.
Remainder Theorem: Let p(x) be any polynomial of degree 1, and a any real number. If p(x) is
divided by x – a. then the remainder is p(a).
Proof: Let us suppose that, when p(x) is divided by x – a, the quotient is q(x) and remainder is r(x).
So we have
p(x) = (x – a) q(x) + r(x), where r(x) = 0 or degree r(x) < degree (x – a)
Since degree of (x – a) is 1, either r(x) = 0 or degree of r(x) = 0 (<1).
So r(x) is a constant, say r.
Thus for all values of x, p(x) = (x – a)q(x) + r where r is a constant.
In particular, for x = a,
p(a) = 0. q(a) + r = 0 + r = r. This proves the theorem.
Illustration 3: What would be the remainder when (4x3 + 7x
2 – 5x + 3) is divided by (x – 1).
Solution: Let f(x) = 4x3 + 7x
2 – 5x + 3.
We may write, (x + 2) = [x – (–2)]
Thus, when f(x) is divided by [x – (–2)], then by remainder theorem,
Illustration 4: Using remainder theorem, find the remainder when (4x3 – 12x
2 + 15x – 3) is divided by (2x – 1).
Solution: Let f(x) = 4x3 – 12x
2 + 15x – 3.
We may write, (2x – 1) = 21
x2
.
Now f 1
2
1 1 1 1 154x 12x 15x 3 3 3 2
8 4 2 2 2
Required remainder = 2.
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If p(x) is a polynomial of degree n > 0, then it follows the remainder theorem that p(x) = (x – a). q(x) + p(a), where q(x) is a polynomial of degree n – 1. If p(a) = 0, then p(x) = (x – a).q(x) , where q(x) is a polynomial of degree n – 1. We say that (x – a) is factor of p(x) Factor Theorem: Let p(x) be polynomial of degree n > 0. If p(a) = 0 for a real number a, then (x – a) is a factor of p(x). Conversely, if (x – a) is a factor of p(x), the remainder p (a) must be zero. Illustration 5: Show that (x + 3) as well as (3x + 2) is a factor of the polynomial (3x
3 + 8x
2 – 5x – 6).
Solution: Let f(x) = 3x3 + 8x
2 – 5x – 6.
By factor theorem, (x + 3) will be a factor of f(x), if f(–3) = 0. Now, f(–3) = 3(–3)
By factor theorem, (3x + 2) will be factor of f(x), if f 2
0.3
Now f
3 22 2 2 2
3x 8x 5x 63 3 3 3
=8 4 10 8 32 10
3x 8x 6 6 0.27 9 3 9 9 3
SQUARE ROOT:
If y x or y2 = x, then y is called the square root of x.
As per the above definition 4 is the square root of 16 since 42 = 16.
But also (–4)2 = 16
The positive square root of 16 is called the principal square root of 16.
In general 2x = x. By square root we mean only the principal square root.
Hence 2x = x when x > 0
2x = x when x < 0
2x = 0 when x = 0.
In short 2x = |x|
In what follows we consider only the non-negative square roots. Modulus sign is to be put whenever it is
needed. For instance 2 2 22 ( ) ( ) ( ( ) ) a ab b a b a b a b if a b .
Illustration 6: Find the square root of a8b
12c
6
Solution:
18 12 6 8 12 6 4 6 32( ) a b c a b c a b c
Illustration 7: Find the square root of
4
2
25
4
a
b
Solution: 4
2
25a
4b=
22 25
2
a
b=
25 a
2 b
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Illustration 8: Find
5
6
121
9
a
ab
Solution: 5
6
121a
9 ba=
22 2
3
11 a
3 b
= 2
3
11a
3b
Square root by inspection:
a2 + 2ab + b
2 = (a + b)
2
a2 – 2ab + b
2 = (a – b)
2
2 2a +2ab+b = (a + b)
2 2a 2ab+b = (a – b)
If the given expression is reduced to one of the above forms, the square root can be written down by
inspection.
Extraction of square root by division method:
Principle behind finding the square root is given below.
Let us find the square root of 25x2 – 30x + 9
Let 25x2 – 30x + 9 = (5x + a)
2
25x2 – 30x + 9 = 25x
2 + 10ax + a
2
–30x + 9 = 10ax + a2
Equating coefficient of x on both sides
10a = –30 or a = –3
We can show this briefly in the division method
5x – 3
5x 25x2 – 30x + 9
25x2 5(x)
2
10x – 3 –30x + 9 (10x – 3) (–3)
–30x + 9
0
Procedure: (1) The square root of 25x
2 is 5x. This is the first term of the square root. Multiply (5x) by (5x) which gives
25x2. Subtract 25x
2 from the dividend. We get –30x + 9 as the next two terms.
(2) Multiply 5x by 2 which gives 10x.30x
10x
gives –3. This becomes the second term of the square root.
Adding –3 to 10x, we get 10x – 3. Multiplying this by –3. This becomes the second term of the square root. Adding –3 to 10x, we get 10x – 3. Multiplying this by –3 we get –30x + 9. The remainder becomes zero.
Square root is (5x – 3)
Square root by the method of Indeterminate coefficients
Illustration 9: Find the square root of x4 + 6x
3 + 17x
2 + 24x + 16
Solution: The expression is of the fourth degree in x. Hence the square root will be an expression of
second degree in x.
x4 + 6x
3 + 17x
2 + 24x + 16 = (x
2 + ax + b)
2
Equating coefficients of the like powers of x on both sides, we get
2a = 6 a = 3
a2 + 2b = 17 2b = 17 – a
2 = 17 – 9 = 8 2b = 8 or b = 4.
Hence the required square root is x2 + 3x + 4.
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Illustration 10: If ax2 + bx + c is a perfect square, then prove that b
2 = 4ac.
Solution: Let ax2 + bx + c = (px + q)
2
ax2 + bx + c = (px + q)
2
Equating coefficients of like powers x
p2 = a
2pq = b 4p2q
2 = b
2
and q2 = c 4ac = b
2
Hence b2 = 4ac
Extraction of square root by division method.
Illustration 11: Find the square root of 4x4 + 12x
3y + 13x
2y
2 + 6xy
3 + y
4
Solution: 2x2 + 3xy + y
2
4x4 +12x
3 y + 13 x
2 y
2 + 6xy
3 + y
4
2x2 4x
4 (2x
2)2 = 4x
4
4x2 + 3xy 12x
3 y + 13 x
2 + y
2 (4x
2 + 3x) 3xy
12x3 y + 9x
2 y
2
2
2
12x ysince 3xy
4x
4x2 + 6xy + y
2 4x
2 y
2 + 6xy
3 + y
4 (4x
2 + 6xy + y
2) y
2
4x2 y
2 + 6xy
3 + y
4
2 22
2
4x ysince y
4x
0
Hence the square root is 2x2 + 3xy + y
2
Illustration 12: Find the square root of x2 – 6x + 17 –
16 2x + 1
2x + 2x + 1
.
Solution: L.C.M. = x2 + 2x + 1
2 2 2 2
2
x x 2x 1 6x x 2x 1 17 x 2x 1 16 2x 1
x 2x 1
4 3 2 3 2 2
2
x 2x x 6x 12x 6x 17x 34x 17 32x 16
x 2x 1
4 3 2
2
x 4x 6x 4x 1
x 2x 1
Numerator:
2
2
2
x
2x 2x
2x 4x 1
4 3 2
4
3 2
3 2
2
2
x 4x 6x 4x 1
x
4x 6x
4x 4x
2x 4x 1
2x 4x 1
0
2x 2x 1
Square root = (x2 – 2x + 1)
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Denominator:
x
2x 1
2
2
x 2x 1
x
2x 12x 1
0
x 1
Square root = (x + 1)
Illustration 13: Find the square root of x2 + 6 +
3 212x 15x +12x 15
4 3 2x 4x + 5x 4x + 4
Solution: Number of the fraction = 12x3 – 15x
2 + 12x – 15 = 12x
3 + 12x – 15x
2 – 15
= 12x (x2 + 1) – 15 (x
2 + 1)
= (x2 + 1) (12x – 15)
Denominator = x4 – 4x
3 + 5x
2 – 4x + 4 = x
3 – 4x
3 + 4x
2 + x
2 – 4x + 4
= x2 (x
2 – 4x + 4) + 1 (x
2 – 4x + 4) = (x
2 – 4x + 4) (x
2 + 1)
Fraction =
2
22 2
x 1 12x 15 12x 15
x 4x 4x 1 x 4x 4
Square root of the given expression = x2 + 6 +
2
12x 15
x 4x 4
2 2 2
2
x x 4x 4 6 x 4x 4 12x 15
x 4x 4
4 3 2 2
2
x 4x 4x 6x 24x 24 24 12x 15
x 4x 4
4 3 2
2
x 4x 10x 12x 9
x 4x 4
Square root of numerator:
2
2
2
x
2x 2x
2x 4x 3
4 3 2
4
3 2
3 2
2
2
x 4x 10x 12x 9
x
4x 10x
4x 4x
6x 12x 9
6x 12x 9
0
2x 2x 3
Square root of Denominator:
x
2x 1
2
2
x 4x 4
x
4x 44x 4
0
x 2
Square root of given expression = 2x 2x 3
x 2
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Homogeneous function
An integral function is said to be homogeneous, if each of its terms is of the same degree with respect to any
set of variables.
Ex. (2x2 + 4y
2 – 5xy) is a homogeneous expression of the second degree in x and y.
The product of two homogeneous functions of degrees m and n respectively is a homogeneous function of
degree m + n.
e.g. (2x + 5y) (4x2 + 5y
2) = 8x
3 + 10 xy
2 + 20x
2y + 25y
3
2x + 5y is of first degree in x and y
4x2 + 5y
2 is of second degree in x and y and both are homogeneous in x and y.
Their product is homogeneous and of third degree in x and y.
The function is said to be symmetric with respect to any set of variables if the interchange of any pairs of the
set of variables does not alter the value of the function.
The function 5x + y + z is symmetric with respect to y and z, but it is not symmetric with respect to x and y.
The expression xy + yz + zx + x + y + z is symmetric with respect to x, y, z.
Similarly 2 2 2 2 2 2
P q r p q rand
qr rp pq p q q r r p
are symmetric with respect to p, q, r.
Homogeneous symmetric integral functions
Degree Variables Functions
1 x, y a(x + y)
2 x, y a(x2 + y
2) + bxy
1 x, y, z a(x + y + z)
2 x, y, z a(x2 + y
2 + z
2) + b (xy + yz + zx)
Illustration 14: Obtain a homogeneous expression of the second degree in x, y, z symmetric with respect to all
the three variables, whose value is 9 when x = y = z =1 and whose value is 3 when x = 0, y = 1,
z = 1.
Solution: Let the homogeneous symmetric expression be
a(x2 + y
2 + z
2) + b(yz + zx + xy)
when x = y = z = 1
3a + 3b = 9 or a + b = 3
when x = 0; y = 1; z = –1
a (0 + 1 + 1) + b (–1 + 0 + 0) = 3
i.e. 2a – b = 3
From (i) and (ii) a + b = 3
2a – b = 3
Adding 3a = 6 or a = 2; b = 3 – a = 3 – 2 = 1
Hence the required expression is a(x2 + y
2 + z
2) + b(yz + zx + xy)
i.e. 2(x2 + y
2 + z
2) + (yz + zx + xy)
Illustration 15: Factorizea, b, c ab (a
2 – b
2)
Solution: f (a) = ab (a2 – b
2) + bc (b
2 – c
2) + ca (c
2 – a
2)
f(b) = b. b (b2 – b
2) + bc (b
2 – c
2) + c . b (c
2 – b
2)
= b3c – bc
3 + bc
3 – b
3c = 0
(a – b) is a factor of the given expression.
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Since the given expression in a cyclic one, the factors will be also cyclic.
They are (b – c); (c – a).
The given expression is of fourth degree.
But the product of the three factors is of third degree.
A linear cyclic factor (i.e. of first degree.) in a, b, c (i.e a + b + c) will be a factor
Let there may be constant factor say K
The given expression can be written as ab (a2 – b
2) + bc (b
2 – c
2) + ca (c
2 – a
2)
K (a – b) (b – c) (c – a) (a + b + c) …….(1)
This being an identity, we can give any values to a, b, c in (1)
Put a = 0; b = 1; c = 2
0.1 (02 – 1
2) + 1.2 (1
2 – 2
2) + 2.0 (2
2 – 0
2)
K (0 – 1) (1 – 2) (2 – 0) (0 + 1 + 2)
2 (1 – 4) = K (–1) (–1) (2) (3) – 6 = K (–1) (–1) (2) (3) –6 = K (6) 6K = –6 K = –1
Factors of the given expression = –1 (a – b) (b – c) 9 (c – a) (a + b + c).
Highest common factor:
We define H.C.F. (Highest Common Factor) or G.C.D.(Greatest Common Divisor) of two polynomials P(x)
and Q(x) to be the common divisor which has the highest degree among all common divisors and which has
the highest degree term coefficient as positive.
Ex. If P(x) = (x – 5)2 (x – 4)
4 (x + 2)
3
Q(x) = (x – 5) (x + 2)2 (x – 4) find the H.C.F
The H.C.F. or G.C.D. of P(x) and Q(x) is (x – 5) (x + 2)3 (x – 4)
If two functions do not have a common divisor they are said to be prime to each other
Rule for finding the H.C.F. by the method of division:
Let f(x) and g(x) be two integral functions and the degree of g(x) be less than that of f(x). Divide f(x) by g(x).
The remainder obtained is the new divisor and g(x) the new dividend. Continue this mode of division, until
there is no remainder. The last divisor will be the required H.C.F.
Note 1: When two given functions are of the same degree, it is immaterial, which of the two given
functions is used as the divisor is the H.C.F. If the remainder reduces to a non- zero constant, then there is
no H.C.F. i.e., the two functions are prime to each other.
Illustration 16: Find the H.C.F. of the functions 10x4 – 7x
3 + x
2, 4x
4 – 2x
3 – 2x + 1
Solution: The first expression is divided by x2, which is not a factor of the second expression. We then
get 10x2 – 7x + 1
Now we will find expression is divided by x2, which is not a factor of the second expression
by 2 and the second by 5.
Thus the expressions 20x2 – 14x + 2 and 20x
4 – 10x
3 – 10x + 5 are obtained
Find the H.C.F. of these two.
20 x2 – 14x + 2 ) 20 x
4 – 10x
3 – 10x + 5 ( x
2
20 x4 – 14x
3 + 2x
2
4x3 – 2x
2 – 10x + 5
Multiply this remainder by 5 and divide by the divisor
20 x2 – 14x + 2 ) 20 x
3 – 10x
2 – 50x + 25 ( 1
20 x3 – 14x
2 + 2x
4x2 – 52x + 25
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Now multiply this remainder by 5 and divide by the divisor
20x2 – 14x + 2 ) 20 x
2 – 260x + 125 ( 1
20x2 – 14x + 2
–246x + 123
Now divide the remainder by –123, we get 2x – 1
2x – 1) 20x2 – 14x + 2 (10x – 2
20x2 – 10x
– 4x + 2
– 4x + 2
0
Hence the H.C.F. is 2x – 1
Least Common Multiple
The least common multiple (L.C.M.) of two or more polynomials is the polynomial of least degree which is
exactly divisible by each of them.
Thus the L.C.M. of x, x3, x
5 is x
5
Illustration 17: Find the L.C.M. of the monomials
(i) 2x3y, 4x
5z, 6y
3z
2
(ii) 4x2yz, 8x
3y
2z
2, 12x
2y
3z
3
Solution: The L.C.M. of 2, 4, 6 is 12
The L.C.M. of x3, x
5 is x
5
The L.C.M. of y, y3 is y
3
The L.C.M. of z, z2 is z
2
L.C.M. of 2x3y, 4x
5y, 6y
3z
2 is 12x
5y
3z
2
The L.C.M. of 4, 8, 12 is 24
The L.C.M. of th part containing variables is x3y
2z
3
Hence required L.C.M. is 24 x3 y
2 z
3
Illustration 18: Find the L.C.M. of the functions 4a2 b(x
2 – 4x + 3), 6b
2 (4x
2 – 4x + 1), 3a(2x
2 – 7x + 3)
Solution: The L.C.M. of 4, 3, 6 is 12
The L.C.M. of a2 b, a and b
2 is a
2 b
2
x2 – 4x + 3 = (x – 1) (x – 3)
2x2 – 7x + 3 = (2x – 1)(x – 3)
4x2 – 4x + 1 = (2x – 1)
2
Hence the required L.C.M. = 12a2b
2 (2x – 1)
2 (x – 1) (x – 3)
Illustration 19: Find the L.C.M. of the integral functions xy(x – y)(y – z), yz(y – z) (z – x), zx(z – x), (x – y)
Solution: The L.C.M. of xy, yz, zx is xyz The L.C.M. of (x – y)(y – z), (y – z)(z – x), (z – x)(x – y) is (x – y)(y – z) (z – x) Hence required L.C.M. is xyz (x – y) (y – z) (z – x)
INVOLUTIONS The process of obtaining algebraic expressions by multiplying a given algebraic expression by itself again and again is called involution. If two algebraic expressions are separated by “=” sign, then we get an algebraic equation. An algebraic equation which is satisfied by all possible values of the variables involved is called an identity. I. (a + b)
2 = a
2 + 2ab + b
2
II. (a b)2 = a
2 2ab + b
2
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III. (a + b + c)2 = a
2 + b
2 + c
2 + 2ab + 2ac + 2bc
IV. (a + b)3 = a
3 + 3 a
2b + 3 ab
2 + b
3
V. (a b)3 = a
3 3a
2b + 3 ab
2 b
3
VI. (x + a) (x a) = x2 a
2
VII. (a + b)2 + (a b)
2 = 2(a
2 + b
2)
VIII. (a + b)2 (a b)
2 = 4ab
Illustration 20: If a + b + c = 6 and ab + bc + ca = 11, find a2 + b
2 + c
2.
Solution: (a + b + c)2 = a
2 + b
2 + c
2 + 2(ab + bc + ca).
a2 = b
2 + c
2 = (a + b + c)
2 2 (ab + bc + ca) = 6
2 (2 11) = 36 22 = 14.
Illustration 21: If ab = 12, bc = 20, ca = 15, find a + b + c.
Solution: ab bc ca = a2 b
2 c
2
We get, a2 b
2 c
2 = 3600 or, abc = 60
Since, ab, bc and ca are all +ve, it implies that, a, b, c are all +ve. Hence, abc is +ve. Hence,
we reject the ve value of abc.
Then a =abc 60
bc 20 = 3,
b = abc 60
ac 15 = 4,
c = abc 60
ab 12 = 5. Thus a + b + c = 12.
Illustration 22: If x +1
x = 4, find x
4 +
1
4x
.
Solution: x2 +
2
1
x =
21
xx
– 2
= 16 – 2
= 14
Thus, x4 +
4
1
x =
2
2
2
1x
x
– 2 = 14
2 – 2 = 194.
A formula is a mathematical statement of a relation between two or more variables. It is generally given as
an equality.
The variable whose value is expressed in terms of the values of the other given variables is called the
subject of the formula.
Illustration 23: Frame a formula for the data – “the final velocity of a body is the sum of its initial velocity and
the product of its acceleration and the time taken for this change in velocity”.
Solution: We first ‘name’ the qualities involved.
Let, the initial velocity = u; the final velocity = v; the acceleration time
i.e., v = u + a t or. v = u + at Note that v, i.e. the final velocity is the subject of the formula in this case.
Linear Equations in Two Variables: The general form of a linear equation in two variables x and y is ax + by + c = 0, a 0, b 0. a, b, c being real numbers.
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To find a solution of a linear equation in two variables, we assign any value to one of the two variables and then determine the value of the other variable from the given equation.
Illustration 24: x – y + 7 = 0.
Solution: Taking x = 0, y = 7
x =1,y = 8 – - - -- - - - - - - - - etc
There are infinite solutions.
Taking any two solutions, we can plot
the corresponding points P and Q. The
line PQ is a graphical representation of
the linear equation.
(7,O)
(O,7) P1
Q
(1,8)
Two properties of the graph are:
1. Every solution of the given equation determines a point on the line PQ.
2. Every point lying on the line PQ satisfies the given equation.
Illustration 25: Plot the graph of the following linear equation. 3x + 2y = 9.
Solution: Get minimum two points and plot. X Y
0
1
2
3
4.5
3
1.5
0
0 1
2
3
4
1
2 3 y
x
x
x
Algebraic Solution of a system of Linear Equations in two variables:
Equations of the form
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
form a system of linear equations in two variables.
Solution of a system of Linear Equations in two variables:
You are already familiar with the three methods
1. Method of elimination by substitution.
2. Elimination method by equating coefficients.
3. Graphical Method.
There is one more method
4. Method of Cross Multiplication (Cramer’s Rule Method)
Let the given system of equations be a1x + b1y = C1
a2x + b2y = C2
Then
1 1
2 2
x
c b
c b
=
1 1
2 2
y
a c
a c
=
1 1
2 2
1
a b
a b
----------(1)
Note: [This is known as Cramer’s method and 1 1
2 2
a b
a b is called determinant,
a1
a2
b1
b2
D=
And its value is a1b2 – a2b1.].
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From (1)
X =
1 1
2 2
1 1
2 2
c b
c b
a b
a b
= x 1 2 1 2
1 2 2 1
D c b b c
D a b a b
y =1 1
2 2
1 1
2 2
a c
a c
a b
a b
= y 1 2 2 1
1 2 2 1
D a c a c
D a b a b
[Dx the determinant obtained by replacing the x – coefficients 1
2
a
a
with
constant terms 1
2
c
c
.Dy the determinant obtained by replacing the coefficients 1
2
b
b
with constant terms 1
2
c
c
.
Method of Solving: First write down the two equations with the constant terms on RHS.
Then find D, Dx and Dy and calculate x & y using the given formula.
Illustration 26: 2x + y = 35 3x + 4y = 65
Solution: 1 1
2 2
a b
a b
2
3
1
4
D=
= 8 – 3 = 5
Dx = 1 1
2 2
c b
c b=
35 1
65 4= 140 – 65 = 75
Dy =1 1
2 2
a c
a c=
2 35
3 65= 130 – 105 = 25
Hence, x = xD 7515
D 5
Y = yD 255
D 5 .
Quadratic Equations
Quadratic Polynomials: A polynomial of degree 2 is called a quadratic polynomial. Given form of quadratic
polynomial is ax2 + bx + c where a, b, c are real numbers such that a 0 and x is a variable.
Zeros or Roots of a Quadratic Polynomial: A real number is called a zero of a quadratic polynomial
P(x) = ax2 + bx + c if P() = 0.
Quadratic Equation: If P(x) is a quadratic polynomial, then P(x) = 0 is called a quadratic equation.
Roots of a Quadratic Equation: Let P(x) = 0 be a quadratic equation, then the zeros of the polynomial P(x)
are called the roots of the equation P(x) = 0, Thus x = is a root of P(x) = 0 if and only if P() = 0. A
quadratic polynomial may or may not have real zeros. A quadratic polynomial can have atmost two zeros or
two real roots.
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Solutions of Quadratic Equations:
1. Factorization Method:
23x 10x 7 3 0
Splitting the middle term
23x 3x 7x 7 3 0
3x(x 3) 7(x 3) 0
(x 3)( 3x 7) 0
Roots are 7
3 and3
2. By Formula Method: Consider the quadratic equation ax
2 + bx + c = 0, a 0.
2 bx cx 0
a a [Dividing throughout by a]
2 b cx x
a a
2 2
2 b b c bx x
a 2a a 2a
[Add
2b
2a
on both sides to make a perfect square].
2 2
2
b b 4acx
2a 4a
2b b 4acx
2a 2a
2b b 4acx
2a 2a
2b b 4ac
2a
(b2 4ac) is called the discriminant of the quadratic equation and denoted by D. The nature of the roots
depends on the value of D.
Nature of the Roots of a Quadratic Equation:
The roots of the equation are real if b2 4ac 0 and complex if b
2 4ac < 0
Further if b2 4ac = 0, then the two roots are equal or coincident each = b/2a.
If D is a perfect square, the roots are distinct and rational.
Summing Up:
D > 0, roots are real and distinct
D = 0, roots are real & coincident
D < 0, roots are complex.
Relation Between the Roots and Coefficients of the Quadratic Equation ax2 + bx + c = 0:
Roots of ax2 + bx + c = 0 are
2b b 4ac
2a 2a
, if the roots are , , then
Sum of Roots = + = b
a
Product of Roots =c
a.
To find equation whose roots are and
(x ) (x ) = 0
x2 – ( + )x + = 0
x2 (sum 0f roots)x + (product of roots) = 0.
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ASSIGNMENTS
SUBJECTIVE
LEVEL–I
1.(i) Find the square root of x2 + 10 +
2
20x 15
x + 4x + 4
.
(ii) Find the square root of (x + 3) (x + 9) (x2 + 12x + 35) + 16
(iii) Find the square root of (x + 2) (x + 4) (x + 6) (x + 8) + 16
(iv) Find the square root of (x + 1) (2x + 1) (3x + 1) (4x + 1) + x4
(v) Find the square root of (a + x) (a + 2x) (a + 3x) (a + 4x) + x4
(vi) What should be added to (x + 2) (x + 4) (x + 6) (x + 8) to make it a perfect square.
2. Show that: (a)
2 2a 2b 2b a
+2b a a 2b
= 4 (b) a
3 + b
3 + c
3 = 3 abc if a + b + c = 0.
(c) x2 +
2
1
x = x
3 +
3
1
x = x
4 +
4
1
x if a + b + c = 0. (d)
2 2 2
b + c c + a a +b+ +
3bc 3ac 3ab= 1 if a + b + c = 0.
3. Find the missing terms of the following perfect squares and identities:
(a) 9a2 + 24ba + ? (b) x
2 – 12x + ? = (x – ?)
2
4. Find the missing terms of the following perfect cubes:
(a) 8a3 + ? + ? + 27b
3 (b) 1 – ? + ? – 8b
3 (c) ? – 12a
2n + ? – n
3 (d) 8x
3 – 24x
2y +? –?
5. Find the coefficient of x2 in the product of (3x
2 + 2x – 4) (x
2 – 3x – 2).
6. If 9x2 – 30x + k is a perfect square, find the value of k.
7. Express (x2 + 8x + 15) (x
2 – 8x + 15) as a difference of two squares.
8. Factorize (a) x – y2 + 3x
2 + 2x – 2y – 3xy (b) xy
3 – y – 3 xy
2 + 3
9. Factorize (i) a2 + b
2 + 2ab + 2bc + 2ca (ii) (x + 2)
2 – 6(x + 2) (y + 5) + 9(y + 5)
2
10. Solve Using Cramer’s Method.
(i) 2x – y – 3 = 0 (ii) x y
+ = a +ba b
4x + y – 3 = 0 2 2
x y+ = 2
a b
(iii) (a – b)x + (a + b)y = 2a2 – 2b
2 (iv) a
2x + b
2y = c
2
(a + b) (x + y) = 4ab b2x + a
2y = d
2
(v) 2 3
+ = 13x y
5 4
= 2 [x 0, y 0]x y
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LEVEL–II
1. The distance from A to B by two different routes are 81 Km and 85 Km. A motor car taking the longer
road travels on the average 2 Km per hour faster than one taking the short route and does the
journey in 15 minutes less. Find the speed of each car.
2. If one root of the equation x2 + px + q = 0 be the squares of the other,
Prove that p3 q(3p1) + q
2 = 0.
3. Construct the equation whose roots are 2 5
2
and
2 5
2
.
4. If a and c are such that the quadratic equation ax2 5x + c = 0 has 10 as the sum of the roots and
also as the product of the roots, find a and c.
5. If and are the roots of the quadratic equation ax2 + bx + c = 0, find the value of
3 +
3 .
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OBJECTIVE
LEVEL–I
1. The homogeneous function of the second degree in x and y having 2x – y as a factor, taking the
value 2 when x = y = 1 and vanishing when x = –1 and y = 1 is
(A) 2x2 + xy – y
2 (B) 3x
2 – 2xy + y
2 (C) x
2 + xy – 2y
2 (D) none of these
2. If f(x) = x2; g(x) = x
3, then the value of
f b f ais
g b g a
(A) 2 2
a b
a ab b
(B)
2 2
a b
a ab b
(C)
2 2
a
a ab b (D) none of these
3. If x2 – 3x + 2 is a factor of the expression x
4 + ax
2 + b, then the value of a and b are given by
(A) a = –5; b = 4 (B) a = 4; b = –5 (C) a = 5; b = –4 (D) none of these
4. If a + b + c = 6; bc + ca + ab = 11; abc = 6, then the value of (1 – a) (1 – b) (1 – c) is
(A) 1 (B) –1 (C) 0 (D) none of these
5. If a + b = 9, x = 5 and a – b – x = 2 then the value of (a – b) [x3 – 2ax
2 + a
2x – (a + b) b
2] is …
(A) 445 (B) 252 (C) 376 (D) none of these
6. Express 4 2
3 2
x 13x 36
x x 6x
in lowest terms.
(A) (x 3)(x 2)
x
(B)
(x 3)(x 2)
x
(C)
(x 3)(x 2)
x
(D)
(x 3)(x 2)
x
7. Express 3 3 3
2 2 2
(x y z 3xyz)
(x y z xy yz zx)
in lowest terms
(A) x + y – z (B) x + y + z (C) x – y + z (D) x – y – z 8. If x
4 – 2x
3 + 3x
2 – mx + 5 is exactly divisible by (x – 3), then m = _______
(A) –4 (B) –40 (C) 40
3
(D) None
9. The value of 6a + 11b, if x
3 – 6x
2 + ax + b is exactly divisible by (x
2 – 3x + 2) is
(A) 0 (B) 132 (C) 66 (D) –66 10. If (x – 1) & (x + 3) are the factors of x
3 + 3x
2 – x – 3 then the other factor is _____
(A) x + 1 (B) x – 1 (C) x – 3 (D) x + 2
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LEVEL–II
1. The sum of x 1
x 1
and its reciprocal is
(A) 2
2
x 1
x 1
(B)
2
2
x 1
x 1
(C)
2
2
2(x 1)
x 1
(D) none
2. If P = 1 2x
1 2x
and Q =
1 2x
1 2x
then
P Q
P Q
= __________.
(A) 2
4x
1 4x (B)
21 4x
4x
(C)
4x
1 4x
(D)
2(1 4x )
4x
3. What should be substracted from2
2
2x 2x 7 x 1toget
x 2x x 6
?
(A) x 2
x 3
(B)
x 2
x 3
(C)
x 2
x 3
(D)
x 2
x 3
4. Simplify: 2 2 2 4
1 1 2x
1 x x 1 x x 1 x x
(A) 0 (B) 2 4
2x
1 2x x+ + (C)
2 4
1
1 x x
(D)
2 4
2x 3
1 x x
5. Simplify:
3 3 32 2 2 2 2 2
3 3 3
a b b c c a
a b b c c a
(A) 3(a + b)(b + c)(c + a) (B) 2(a + b)(b + c)(c + a)
(C) (a + b)(b + c)(c + a) (D) 1
6. Sum of the roots of the equation 4x2 – 1= 0
(A) 1
4 (B) 0 (C)
1
4 (D) 4
7. Frame an equation in x having 2 and –3 as roots
(A) x2 + x + 6 = 0 (B) x
2 – x – 6 = 0 (C) x
2 – 6 = 0 (D) x
2 + x – 6 = 0
8. The roots of (x2 – 3x + 2)(x)(x – 4) = 0 are
(A) 4 (B) 0 and 4 (C) 0, 1, 2 and 4 (D) 1, 2 and 4
9. The difference of the roots of x2 – 7x – 9 = 0 is
